Question Video: Writing the Absolute Value Inequality That Represents a Given Solution Set | Nagwa Question Video: Writing the Absolute Value Inequality That Represents a Given Solution Set | Nagwa

Question Video: Writing the Absolute Value Inequality That Represents a Given Solution Set Mathematics

Determine the absolute value inequality representing π‘₯ ∈ ℝ βˆ’ [βˆ’21, 27].

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Video Transcript

Determine the absolute value inequality representing π‘₯ is the set of all real numbers minus the closed interval negative 21 to 27.

First, let’s consider the information we know about π‘₯. If we think about a number line showing all real numbers, and then we add negative 21 and 27, these are all the excluded values of π‘₯, the values that π‘₯ cannot be. Let’s add the values that π‘₯ can be to the same number line. π‘₯ cannot be equal to 27, but it can be greater than 27. And then π‘₯ cannot be equal to negative 21, but it can be less than that, which means π‘₯ is less than negative 21.

But now let’s think about what we know about absolute value. The definition of absolute value tells us that it is the magnitude of a number without regard to its sign. If we have the values of negative π‘₯ and π‘₯ on the number line, we say that the absolute value of π‘₯ is π‘₯. And both values, negative π‘₯ and π‘₯, are π‘₯ units from zero; they have a magnitude of π‘₯.

And to write an inequality to represent this value of π‘₯, what we’ll want to do is find the middle of negative 21 and 27. If we find the midpoint, the distance between negative 21 and π‘š will be the same as the distance from π‘š to 27. The distance from negative 21 to 27 is 48 units. Half of that is 24. 24 units to the right of negative 21 on the number line is three. And three plus 24 equals 27. And what we’re saying here is that π‘₯ must be more than 24 units from three. If we start at three, we need to go more than 24 units to get into the range of what π‘₯ can be in the right direction. And if we start at three, we need to go more than 24 units in the left direction to find a value that is acceptable for π‘₯.

And now we need to translate that into an absolute value inequality. We can write that as the absolute value of π‘₯ minus three must be greater than 24. To check that this is true, we break it up into two separate equations, which is π‘₯ minus three is greater than 24 and the negative of π‘₯ minus three is greater than 24. On the left, we add three to both sides to get π‘₯ is greater than 27. On the right, we multiply both sides of the inequality by negative one so that π‘₯ minus three is less than negative 24.

And after we add three to both sides, we get that π‘₯ is less than negative 21 which confirms the absolute value of π‘₯ minus three must be greater than 24. And that will produce π‘₯-values such that π‘₯ can be all reals with the exception of the closed interval negative 21 to 27.

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